SOLUTION: Find the radius of a circle defined by the equation 6x^2+6y^2+48x+108y+558=0 They want you put it in this form (x-h)^2+(y-k)^2=r^2 I first Subtracted the 558 on both sides but I

Algebra ->  Circles -> SOLUTION: Find the radius of a circle defined by the equation 6x^2+6y^2+48x+108y+558=0 They want you put it in this form (x-h)^2+(y-k)^2=r^2 I first Subtracted the 558 on both sides but I       Log On


   

Question 372491: Find the radius of a circle defined by the equation 6x^2+6y^2+48x+108y+558=0
They want you put it in this form (x-h)^2+(y-k)^2=r^2
I first Subtracted the 558 on both sides but I got stuck after that
Here's what I have
6x^2+6y^2+48x+108y+558-558=0-558
6x^2+6y^2+48x+108y=558
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
You need to complete the square in x and y.
6x%5E2%2B48x%2B6y%5E2%2B108y%2B558=0
x%5E2%2B8x%2By%5E2%2B18y%2B93=0
%28x%5E2%2B8x%2B16%29%2B%28y%5E2%2B18y%2B81%29%2B93=16%2B81
%28x%2B4%29%5E2%2B%28y%2B9%29%5E2%2B93=97
%28x%2B4%29%5E2%2B%28y%2B9%29%5E2=4
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Center:(-4,-9)
Radius: 2